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Question Number 93109 by Mikael_786 last updated on 10/May/20

∫_0 ^(2π)  cos^(2020) (x)dx

$$\underset{\mathrm{0}} {\overset{\mathrm{2}\pi} {\int}}\:{cos}^{\mathrm{2020}} \left({x}\right){dx} \\ $$

Commented by prakash jain last updated on 11/May/20

∫_9 ^(2π) cos^(2020) xdx  ∫_0 ^π cos^(2020) xdx+∫_π ^(2π) cos^(2020) xdx          ∫_π ^(2π) cos^(2020) xdx=∫_0 ^π cos^(2020) (π+u)du           =∫_0 ^π cos^(2020) udu  ∫_9 ^(2π) cos^(2020) xdx=2∫_0 ^π cos^(2020) xdx  =2(∫_0 ^(π/2) cos^(2020) xdx+∫_(π/2) ^π cos^(2020) xdx)       substitute x=(π/2)+u to get  =4∫_0 ^( π/2) cos^(2020) xdx  =4×((2019.2017.2015...1)/(2020∙2018∙2016....2))× (π/2)

$$\int_{\mathrm{9}} ^{\mathrm{2}\pi} \mathrm{cos}^{\mathrm{2020}} {xdx} \\ $$$$\int_{\mathrm{0}} ^{\pi} \mathrm{cos}^{\mathrm{2020}} {xdx}+\int_{\pi} ^{\mathrm{2}\pi} \mathrm{cos}^{\mathrm{2020}} {xdx} \\ $$$$\:\:\:\:\:\:\:\:\int_{\pi} ^{\mathrm{2}\pi} \mathrm{cos}^{\mathrm{2020}} {xdx}=\int_{\mathrm{0}} ^{\pi} \mathrm{cos}^{\mathrm{2020}} \left(\pi+{u}\right){du} \\ $$$$\:\:\:\:\:\:\:\:\:=\int_{\mathrm{0}} ^{\pi} \mathrm{cos}^{\mathrm{2020}} {udu} \\ $$$$\int_{\mathrm{9}} ^{\mathrm{2}\pi} \mathrm{cos}^{\mathrm{2020}} {xdx}=\mathrm{2}\int_{\mathrm{0}} ^{\pi} \mathrm{cos}^{\mathrm{2020}} {xdx} \\ $$$$=\mathrm{2}\left(\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{cos}^{\mathrm{2020}} {xdx}+\int_{\pi/\mathrm{2}} ^{\pi} \mathrm{cos}^{\mathrm{2020}} {xdx}\right) \\ $$$$\:\:\:\:\:\mathrm{substitute}\:{x}=\frac{\pi}{\mathrm{2}}+{u}\:\mathrm{to}\:\mathrm{get} \\ $$$$=\mathrm{4}\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} \mathrm{cos}^{\mathrm{2020}} {xdx} \\ $$$$=\mathrm{4}×\frac{\mathrm{2019}.\mathrm{2017}.\mathrm{2015}...\mathrm{1}}{\mathrm{2020}\centerdot\mathrm{2018}\centerdot\mathrm{2016}....\mathrm{2}}×\:\frac{\pi}{\mathrm{2}} \\ $$

Commented by Mikael_786 last updated on 10/May/20

thank u Sir

$${thank}\:{u}\:{Sir} \\ $$

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